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A081090
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a(n)^2 + 1 = A081089(n), where A081089(n) = A081088(n+1)/A081088(n); involves the partial quotients of a series of continued fractions that sum to unity.
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4
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OFFSET
| 1,2
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COMMENTS
| log(a(n+1))/log(a(n)) -> 1+sqrt(2). The 8-th term has 59 digits, while the 9-th term has 141 digits.
sum(n>0, a(n)/a(n+1) ) = 1/2; the ratio of the terms a(n+1)/a(n), for n>1, form the convergents of the continued fraction series described by A081088; thus a(n+1) = A081088(n)*a(n) + a(n-1), for n>1. - Paul D. Hanna (pauldhanna(AT)juno.com), Mar 05 2003
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FORMULA
| a(n) = a(n-2)*(a(n-1)^2 + 1) for n>3, with a(1)=0, a(2)=2, a(3)=5. Also, a(n)*a(n-1) = A081088(n) for n>2, a(n) = a(n-2)*A081089(n-1) for n>2.
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CROSSREFS
| Cf. A081086, A081088, A081089.
Sequence in context: A102011 A004098 A005114 * A071880 A071882 A206848
Adjacent sequences: A081087 A081088 A081089 * A081091 A081092 A081093
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KEYWORD
| nonn,easy
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AUTHOR
| Hans Havermann (gladhobo(AT)teksavvy.com) and Paul D. Hanna (pauldhanna(AT)juno.com), Mar 05 2003
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